Problem: Divide the following complex numbers. $ \dfrac{8+i}{-2+i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2-i}$ $ \dfrac{8+i}{-2+i} = \dfrac{8+i}{-2+i} \cdot \dfrac{{-2-i}}{{-2-i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(8+i) \cdot (-2-i)} {(-2+i) \cdot (-2-i)} = \dfrac{(8+i) \cdot (-2-i)} {(-2)^2 - (1i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(8+i) \cdot (-2-i)} {(-2)^2 - (1i)^2} = $ $ \dfrac{(8+i) \cdot (-2-i)} {4 + 1} = $ $ \dfrac{(8+i) \cdot (-2-i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({8+i}) \cdot ({-2-i})} {5} = $ $ \dfrac{{8} \cdot {(-2)} + {1} \cdot {(-2) i} + {8} \cdot {-1 i} + {1} \cdot {-1 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-16 - 2i - 8i - 1 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-16 - 2i - 8i + 1} {5} = \dfrac{-15 - 10i} {5} = -3-2i $